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 June 21st, 2014, 05:15 PM #1 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 Which side is greater? my challenge #2 $\displaystyle \sqrt{2} \ + \ \sqrt{3} \ + \ \sqrt{5} \ + \ \sqrt{7} \ \ \ versus \ \ \ 8$ I am not asking to do the actual work to determine which side is larger. The question is: At any given step, is it possible to eliminate one of the four radicals that will be present in that step after squaring, adding, and/or subtracting? If that can be done, then the original problem is possible by my method illustrated here: Which side is greater? my challenge #1 post #3 Last edited by Math Message Board tutor; June 21st, 2014 at 05:19 PM.
 June 21st, 2014, 06:08 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra If you get two roots only on one side and then square, they will become one after squaring.
June 22nd, 2014, 07:45 AM   #3
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Quote:
 Originally Posted by v8archie If you get two roots only on one side and then square, they will become one after squaring.
Yes, as in

$\displaystyle \sqrt{a} \ + \ \sqrt{b} \ = \ something$

or

$\displaystyle \sqrt{a} \ - \ \sqrt{b} \ = \ something$

June 22nd, 2014, 08:51 AM   #4
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Quote:
 Originally Posted by v8archie If you get two roots only on one side and then square, they will become one after squaring.
That doesn't automatically solve the problem. If you transpose
$$\sqrt2+\sqrt3>^?8-\sqrt5-\sqrt7$$
and square you do have only one radical on the left, but you have three on the right:
$$5+2\sqrt6>^?76-16\sqrt5-16\sqrt7+2\sqrt{35}$$

Now fortunately that's not a problem here since you can transpose and square again to get
$$24>^?8253-2720\sqrt5-2592\sqrt7+796\sqrt{35}$$
then move the negatives around (and fold the constant) to get
$$2720\sqrt5+2592\sqrt7>^?8229+796\sqrt{35}$$
which can be squared to get one radical on each side and solved by (transposing and squaring) twice. But this isn't general.

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